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  • Let PQ and RS be tangents at the extremities of the diameter PR of a circle of radius r. If PS and RQ intersect at point x on the circumference of the circle, then 2r 

Let PQ and RS be tangents at the extremities of the diameter PR of a circle of radius r. If PS and RQ intersect at point x on the circumference of the circle, then 2r equals

Option: 1

\mathrm{\sqrt{P Q \cdot R S}}


Option: 2

\mathrm{\frac{P Q+R S}{2}}


Option: 3

\mathrm{\frac{2 P Q+R S}{P Q+R S}}


Option: 4

\mathrm{\sqrt{\frac{P Q^2+R S^2}{2}}}


Answers (1)

best_answer

As the triangle PQR is a right triangle,

\mathrm{\begin{aligned} & \therefore \frac{P Q}{P R}=\cot \theta \\ & \Rightarrow \frac{P Q}{2 r}=\cot \theta \end{aligned}}     ...[1]

Also, as ?PRS is a right triangled, therefore

\mathrm{\begin{aligned} & \frac{R S}{P R}=\cot \left(\frac{\pi}{2}-\theta\right) \\ & \Rightarrow \frac{R S}{2 r}=\tan \theta \end{aligned}}     ...[2]

From (1) and (2), we get

\mathrm{\begin{aligned} & \left(\frac{P Q}{2 r}\right)\left(\frac{R S}{2 r}\right)=\cot \theta \cdot \tan \theta=1 \\ & \Rightarrow(P Q)(R S)=(2 r)^2 \Rightarrow 2 r=\sqrt{(P Q) \cdot(R S)} \end{aligned}}

Posted by

Devendra Khairwa

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